Question

in a polynomial with degree 3, can we find a simple expression for the sum of the cubes of the three roots?

Answer 1

In polynomial ax^3+bx^2+cx+d = 0
3 roots are p, q and r
p+q+r = -b/a
pq+pr+qr = c/a
pqr = -d/a

can we find a simple expression for p^3+q^3+r^3 using the above identities?

Answer 2

For example, in polynomial ax^2+bx+c = 0
roots are p and q

p+q = -b/a
pq = c/a

p^2+q^2 = (p+q)^2 - 2pq

Answer 3

eh?

Answer 4

I doubt it. Do you know how to solvethe cubic?

Answer 5

what do you mean by solve the cubic?

Answer 6

\[\begin{align}
(p+q+r)^3&=p^3+q^3+r^3+3p^2q+3p^2r+3pq^2+6pqr+3pr^2+3q^2r+3qr^2\\
&=p^3+q^3+r^3+3(p^2q+p^2r+pq^2+2pqr+pr^2+q^2r+qr^2)\\
&=p^3+q^3+r^3+3(p(pq+pr+qr)+q(pq+pr+qr)+pr^2+qr^2)\\
&=p^3+q^3+r^3+3((p+q)(pq+pr+qr)+pr^2+qr^2)\\
&=p^3+q^3+r^3+3((p+q+r)(pq+pr+qr)-r(pq+pr+qr)+pr^2+qr^2)\\
&=p^3+q^3+r^3+3((p+q+r)(pq+pr+qr)-pqr)
\end{align}\]therefore:\[p^3+q^3+r^3=(p+q+r)^3-3((p+q+r)(pq+pr+qr)-pqr)\]
I think that is correct.

Answer 7

Solve the cubic equation for its 3 roots.

Answer 8

@inkyvoyd, that's what we're trying to avoid

Answer 9

Thanks heaps, this is now closed

Answer 10

yw

Answer 11

@psujono , medal asnaseer if you think answer was good!

Answer 12

thx for the thought inkyvoyd, but I use this site to learn and teach rather than gain any medals, so I am not really bothered if I don't get a medal. I just enjoy the interaction with other users. :)

Answer 13

I'll medal you anyway, good answer - thanks!

Answer 14

thx

Answer 15

@asnaseer , there are people like me that prefer medals (LOL), which is why I reminded him to medal (so, you don't forget xD, because if someone like me helps you out, the will rage without a medal)

Answer 16

@inkyvoyd - Good point - I guess I should think of everyone else using this site as well :)

Answer 17

No, just me LOL